Otherwise, given a fixed $(p,k)$ once you have a nontrivial solution you get infinitely many
using automorphs of the indefinite part in variables $(c,d).$ That is, there may be many parametrized families of solutions of one type or another. But you can figure some of those out with a computer algebra system more easily than I can by hand.

The next interesting case is when $12 k^2 + 8p$ is a square, which means that the binary form
$T(c,d)=2c^2+2kcd-(p+k^2)d^2$ factors. So $3 k^2 + 2p$ is a square, which is not possible for even $k,$ so $k$ is odd and $2p \equiv 6 \pmod 8,$ or $p \equiv -1 \pmod 4.$ Unless $p=3$ we also need
$p \equiv -1 \pmod 3,$ or $p \equiv -1 \equiv 11 \pmod {12}.$

For what it's worth, you can write the equation in the form
$$ a^2 + (c-kd)^2 + pb^2 + pd^2 = 3c^2, $$
so you are looking at a parametrized subset of the equation
$$ A^2 + B^2 + pC^2 + pD^2 = 3c^2. $$
If $p \equiv 1 \bmod 4$, then $p(B^2+D^2) = R^2 + S^2$ is a sum of two squares, and your solutions must occur among those of
$$ R^2 + S^2 + T^2 + U^2 = 3c^2. $$
Both quadrics can be parametrized by the standard method of sweeping lines if you know
one solution. For arbitrary primes $p$ such a solution seems to be difficult to find. And even armed with such a parametrization you then would have to figure out which of them satisfy the additional conditions coming from the original equation.